Finding Leg Lengths. Find the length of a leg in the 45°-45°-90°
triangle.

Answer:
12. [tex]x=10[/tex]
13. [tex]x=4[/tex]
14. [tex]x=5[/tex]
15. [tex]x=1[/tex]
16. [tex]x=8[/tex]
17. [tex]x=14[/tex]
Step-by-step explanation:
for these i also used the pythagorean theorem ([tex]a^{2}+b^{2}=c^{2}[/tex])
[tex]a[/tex] and [tex]b[/tex] are the legs, [tex]c[/tex] is the hypotenuse
12.
[tex]x^{2} +x^{2} =(10\sqrt{2} )^{2}[/tex]
[tex]2x^{2} =200[/tex]
[tex]x^{2} =100[/tex]
[tex]x=10[/tex]
13.
[tex]x^{2} +x^{2} =(4\sqrt{2} )^{2}[/tex]
[tex]2x^{2} =32[/tex]
[tex]x^{2} =16[/tex]
[tex]x=4[/tex]
14.
[tex]x^{2} +x^{2} =(5\sqrt{2} )^{2}[/tex]
[tex]2x^{2} =50[/tex]
[tex]x^{2} =25[/tex]
[tex]x=5[/tex]
15.
[tex]x^{2} +x^{2} =(\sqrt{2} )^{2}[/tex]
[tex]2x^{2} =2[/tex]
[tex]x^{2} =1[/tex]
[tex]x=1[/tex]
16.
[tex]x^{2} +x^{2} =(8\sqrt{2} )^{2}[/tex]
[tex]2x^{2} =128[/tex]
[tex]x^{2} =64[/tex]
[tex]x=8[/tex]
17.
[tex]x^{2} +x^{2} =(14\sqrt{2} )^{2}[/tex]
[tex]2x^{2} =392[/tex]
[tex]x^{2} =196[/tex]
[tex]x=14[/tex]